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# Homework Portfolio

1.

- 1.71. Our proof of the Cauchy-Schwarz inequality, Theorem 1.13, used that when $$U$$ is a unit vector, $0 \leq ||V−(U·V)U||^2 = ||V||2 −(U·V)^2$. Therefore if $$U$$ is a unit vector and equality holds, then $$V = (U · V)U$$. Show that equality occurs in the Cauchy Schwarz inequality for two arbitrary vectors $$V$$ and $$W$$ only if one of the vectors is a multiple (perhaps zero) of the other vector.

Answer: In the first case, when $$W=0$$, the $$W$$ is a multiple of $$V$$; In the second case, when $$W$$ is nonzero, then consider the unit vector $$U = \frac{W}{||W||}$$. Then, by the result in the question, it follows that $$V = (U · V)U$$. Therefore:

$V = (U · V)U = (\frac{W}{||W||}· V)·\frac{W}{||W||}= (\frac{W·V}{||W||^2})·W$ As $$(\frac{W·V}{||W||^2})$$ is a constant, so $$V$$ is a multiple of $$W$$.

2.

-2.19. Suppose C is an n by n matrix with orthonormal columns. Use Theorem 2.2 to show that $||CX|| \leq \sqrt{n} ||X||$ Use the Pythagorean theorem and the result of Problem 2.17 to show that in fact $||CX|| = ||X||$ for such a matrix.

Answer: (1) First we calculate $$C$$. Let $$C_j$$ denote the $$j$$th column of C. Since C has orthonormal columns, each $$C_j$$ has norm 1. Then $||C|| = \sqrt{\mathop{\sum_{i=1}^n\sum_{j=1}^n}C_{ij}^2}$ $=\sqrt{\mathop{\sum_{j=1}^n(\sum_{i=1}^n}C_{ij}^2)}$ $=\sqrt{\sum_{j=1}^n|| C_j||^2}$ Now by Theorem 2.2, $||CX|| \leq ||C||||X|| = \sqrt{n}||X||$ as desired. (2)By Problem 2.17, $$CX=x_1C_1 +x_2C_2 +\dots+x_nC_n$$. To find the norm of RHS, we need to apply the Pythagorean theorem (we need that C has orthonormal columns) to get $||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2 = ||x_1 C_1||^2 + ||x_2 C_2||^2 + \dots + ||x_n C_n||^2$ Now we can put the pieces together: $||CX||^2 = ||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2$ $=||x_1 C_1||^2 + ||x_2 C_2||^2+\dots+ ||x_n C_n||^n$ $=x_1^2||C_1||^2 + x_2^2|| C_2||^2+\dots+ x_n ^2||C_n||^n$ $=x_1^2 + x_2^2+\dots+ x_n ^2$ $= ||X||^2$ Since norms are nonnegative, we can conclude that $$||CX|| = ||X||$$.

3.

-2.44. Use the Cauchy-Schwarz inequality $|A · B| \leq||A||||B||$

to prove: (a) the function $$f (X) = C · X$$is uniformly continuous,

(b) the function $$g(X, Y) = X · Y$$ is continuous.

Answer: (a) In case 1, If $$C=0$$ then $$f(X)=0$$ for all $$X$$,so $$|f(X)−f(Y)|=0<ε$$ for all $$ε,X,Y$$.

In case 2, where $$C=(c1,c2,...,cn)\neq(0,0,...,0)$$.

By the definition of f and properties of the dot product, $$|f(X)− f(Y)|=|C·Y−C·Y|=|C·(X−Y)|$$.

By the Cauchy-Schwartz inequality we get $|f(X)− f(Y)|=|C·(X−Y)|\leq||C||||X−Y||$ Let $$ε>0$$ and take $$δ=\frac {ε}{||C||}$$

If $$||X−Y||<δ= \frac {ε}{||C||}$$ Then we have $$|f(X)− f(Y)|\leq||C||||X−Y||<ε$$ for all $$X$$ and $$Y$$ in $$Rn$$. Therefore for any C, f is uniformly continuous.

(b) Fix $$V (A,B)$$ in $$R^{2n}$$, to show that $$g(V)= g(X,Y)= XY$$ is continuous at $$(a,b)$$

Given $$\epsilon>0$$, we need to find $$\delta>o$$ If $$||U-V||=\sqrt {(X-A)^2+(Y-B)^2}<\delta$$

then $$||g(U)-g(V)||=|XY-AB|<\epsilon$$

By the triangle inequality, we know that $|XY-AB|= |[(X-A)+A][(Y-B)+B]-AB|$ $=|(X-A)(Y-B)+B(X-A)+A(Y-B)|$ $\leq||X-A||+||B|| ||X-A||+||A|| ||Y-B||$ If $$\sqrt {(X-A)^2+(Y-B)^2}<\delta$$ Then $$||X-A||+||B|| ||X-A||+||A|| ||Y-B||$$ $\leq (1+||A||+||B||)\sqrt {(X-A)^2+(Y-B)^2}$ $\leq(1+||A||+||B||)\epsilon$ $\leq(1+||A||+||B||) \frac {\delta}{1+||A||+||B||}$ $=\delta$ So, given $$\epsilon>0$$, set $$\delta= min(1,\frac {\delta}{1+||A||+||B||})$$ we have $$\sqrt {(X-A)^2+(Y-B)^2}<\delta$$ Therefore, for any $$(A,B)$$ in $$R^{2n}$$, $$g(X, Y) = X · Y$$ is continuous.

4.

-2.45. In the triangle inequality $$||A + B|| \leq ||A|| + ||B||$$ put $$A = X − Y$$ and $$B = Y$$. Deduce $$||Y|| − ||X|| \leq ||Y − X||$$. Show that if two points are within one unit distance of each other, then the difference of their norms is less than or equal to one.

Answer: Let $$A$$ and $$B$$be in $$R_n$$. Apply the triangle inequality we have $||A + B|| \leq ||A|| + ||B||.$ Let$$X=A+B$$, $$Y=B$$,so $$A=X−Y$$. so we have $||X|| \leq ||X − Y|| + ||Y||, ||X|| − ||Y|| \leq ||X − Y||.$

Exchange the symbol $$X$$ and $$Y$$ we get $||Y|| − ||X|| \leq ||Y − X||.$ So when $$X$$ and $$Y$$ are within one unit of distance of each other which means $$||Y − X|| \leq 1$$, by the inequality we can conclude $||Y|| − ||X|| \leq ||Y − X||\leq 1$

5.

-3.9. Suppose a function $$F$$ from $$R_n$$ to $$R_m$$ is differentiable at $$A$$. Justify the following statements that prove $L_AH = DF(A)H,$ that is, the linear function LA in Definition 3.4 is given by the matrix of partial derivatives$$DF(A).$$

(a) There is a matrix$$C$$ such that $$L_A(H) = CH$$ for all $$H$$.

(b) Denote by $$C_i$$ the $$i-th$$ row of $$C$$. The fraction $\frac{||F(A+H)−F(A)−L_A(H)||}{||H||} = \frac{||F(A+H)−F(A)−CH||}{||H||}$ tends to zero as $$||H||$$ tends to zero if and only if each component $\frac {f_i(A+H)− f_i(A)−C_iH}{ ||H||}$ tends to zero as $$||H||$$ tends to zero.

(c) Set $$H = h e_j$$ in the $$i-th$$component of the numerator to show that the partial derivative $$f_{i,x_j} (A)$$ exists and is equal to the $$(i, j)$$ entry of $$C$$.

Answer: (a) $$F$$ from $$R_n$$ to $$R_m$$ is differentiable at $$A$$. By the definition of differentiability there is a linear function $$L_A(H)$$ such that $\frac{||F(A+H)−F(A)−L_A(H)||}{||H||}$

tends to $$0$$ as $$||H||$$ tends to zero.

By Theorem 2.1, every linear function from $$R_n$$ to $$R_m$$ can be written as $$LA(H)=CH$$ for all $$H$$, where $$C$$ is some $$m × n$$ matrix.

(b) The absolute value of each component of a vector is less than or equal to the norm of the vector so for each H and i we have $0 \leq| f_i(A + H) − f_i(A) − C_i · H| \leq ||F(A + H) − F(A) − CH||$ where $$Ci$$ is the $$i-th$$ row of $$C$$.

Since $\frac{||F(A+H)−F(A)=LA(H)||}{||H||}$ tends to 0,

by the squeeze theorem, both$$\frac{ | f_i(A+H)− f_i(A)−C_i·H|}{||H|| }$$ and$$\frac{ f_i(A+H)− f_i(A)−C_i·H}{||H|| }$$

(c) Let $$H = he_j = (0,0,...,1,0,...,0)$$, the 1 in the$$j-th$$ place. By part (b), tend to zero as ||H|| tends to zero. $\lim_{\||H||=|h|\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{|h|} =0$

Therefore $\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)−hC_i ·e_j}{h} =0$

Since $$C_i · e_j = c_{i j}$$ and by the definition of partial derivatives,we have $\lim_{h\to\ 0 } \frac{f_i(A+he_j)− f_i(A)}{h} = \frac {\partial f_i}{\partial x_j} (A)$

So we can conclude that $$c_{i j}= c (A)$$

6.

6.14. Justify the following items which prove:

If f is continuous on $$R_2$$ and $$\int_{R}f dA = 0$$ for all smoothly bounded sets $$R$$, then  is identically zero.

(a) If $$f(a,b)=p>0$$ then there is a disc $$D$$ of radius $$r>0$$ centered at $$(a,b)$$in which $$f(x,y)> \frac {1}{2}p$$

(b) If f is continuous and $$f(x,y) ≥ p_1 > 0$$ on a disk $$R$$ then $$\int_{R}f dA \geq p_1(Area(R))$$. $$\int_{R}f dA = 0$$for all smoothly bounded regions $$R$$, then $$f$$ cannot be positive at any point. (d) $$f$$ is not negative at any point either. (e) $$f = 0$$ at all points.

Answer: First, we can assume that f is continuous on R2 and $$\int_{R}f dA = 0$$ for all smoothly bounded sets $$R$$

(a) Because f is continuous at $$(a, b)$$, by the definition of continuity, there is $$r > 0$$ such that for all $$(x, y)$$ such that$$||(x, y) − (a, b)|| < r$$, we have $$|f(x, y) − f(a, b)| < p/2$$.Then we assume p > 0, so $$p − p/2 < f(x,y) < p + p/2$$ In particular, $$f (x, y) > p/2$$

(b) As $$R$$ is bounded, the closure of $$R$$ is closed and bounded. So we can apply the extreme value theorem which means $$f$$ is bounded on the closure of $$R$$. In particular, $$f$$ is bounded on $$R$$. $$f$$ is also integrable on $$R$$; in fact $$\int_{R}f dA = 0$$. Apply the lower bound property, $$\int_{R}f dA \geq p_1(Area(R))$$ holds.

(c) Suppose $$f$$ is positive at $$(a, b)$$.

From $$(a)$$, there is a disc $$R$$of nonzero radius on which $$f(x,y) > f(a,b)/2 > 0$$.

From (b), $$\int_{R}f dA \geq (f(a,b)/2)·area(R) > 0$$ But we assumed that $$\int_{R}f dA = 0$$ for all smoothly bounded sets $$R$$, it comes to a contradiction. Therefore $$f$$ cannot be positive at any point.

(d) As we know that $$−f$$ is continuous, and that for all smoothly bounded regions $$R$$, by linearity, we have $$−fdA=− fdA=−0=0$$ . From(c),we know that $$-f$$ cannot be positive at any point. Thus, we conclude that f cannot be negative at any point.

(e) Therefore, for any $$(a,b)$$, $$f(a,b)$$ is defined and is neither positive nor negative, so it must be $$0$$.

7.

-6.44. Justify the following steps to prove that if $$f$$ is integrable on $$R_2$$ and $$g$$ is a continuous function with $$0 \leq g \leq f$$ then $$g$$ is integrable on $$R_2$$.

(a) $$\int_{D(n)} g dA$$ exsits

(b) $$0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA$$

(c) The numbers $$\int_{D(n)} g dA$$ are an increasing sequence bounded above.

(d) $$\lim_{n\to\infty} \int_{D(n)} g dA$$ exsits

Check $$D : D= R^2$$ unbounded, g 0, continuous, so we need to prove $$\lim_{n\to\infty} \int_{D(n)} g dA$$ exsits.

(a) $$g \geq 0$$ is continuous on $$R_2$$ and $$D(n)$$ is bounded for each $$n$$ so $$g$$ is integrable over $$D(n)$$

(b) By theorem 6.9 $$L area(D) \leq I (f, D)$$ and the fact 0g, we know that $$0 area(D)\leq \int_{D(n)} g dA$$ so if $$0 \leq f(x,y)−g(x,y)$$ then $0=0 area(D)\leq \int_{D(n)} f(x,y)-g(x,y) dA \leq \int_{D(n)} f dA - \int_{D(n)} g dA$ Therefore, $$0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA$$

(c) Let $$C_n =\int_{D(n)} g dA$$. Because $$g \geq 0$$, $$D(n) \leq D(n+1)$$. Then $$C_1, C_2, C_3...C_n$$ is an increasing sequence. Since $$0 \leq \int_{D(n)} g dA \leq \int_{D(n)} f dA$$ and $\lim_{n\to\infty} \int_{D(n)} f dA = \int_{R ^2} f dA$ exists, We got $\int_{D(n)} g dA \leq \int_{R ^2} f dA$

(e) By the Monotone Convergence Theorem for sequences, $$\int_{D(n)} g dA$$ increasing and bounded above is convergent so $$\lim_{n\to\infty} \int_{D(n)} g dA = \lim_{n\to\infty} C_n$$ exists

8.

6.50. Justify steps (a)–(d) to prove that if a continuous function $$f$$ is integrable on an unbounded set $$D$$ then $$\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$$

(a)$$\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} \left|f \right|dA$$

(b)$$\int_ {D} (-f) dA \leq \int_ {D} \left|f \right|dA$$

(c)$$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$

(d)$$\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$$

(a) By Definition 6.9, if $$f$$ is continuous and integrable on an unbounded set $$D$$, then $$| f |$$ is integrable on $$D$$. Rewrite $$f(x,y) = f_+(x,y)− f_−(x,y)$$ where $$f_+(x,y) = f(x,y)$$ if $$f(x,y) \geq 0$$ and 0 otherwise, and $$f_−(x,y) = −f(x,y)$$if $$f(x,y)\leq 0$$ and $$0$$ otherwise. So, by the definition of $$\int_ {D} f dA$$, $\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA$ Since $$\int_ {D} f_- dA$$ is nonnegtive $\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA$ Since $$f_+ ≥ 0$$ and $$f_− ≥ 0$$ are integrable over $$D$$ $\int_ {D(n)} f_+ dA + \int_ {D(n)} f_- dA =\int_ {D(n)} (f_+ + f_-) dA$ By the properties of limits of increasing sequence D(n), we know $$\int_ {D(n)} (f_+ + f_-) dA$$ converges so $\int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} (f_+ + f_-) dA$ By the equation $$f(x,y) = f_+(x,y)− f_−(x,y)$$, we got $\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$

(b) In the same way, we apply (a) to the functions − f to get $\int_ {D} -f dA \leq \int_ {D} \left|-f \right|dA= \int_ {D} \left|f \right|dA$

(c)By the properties of limits and the equation $$\int_ {D(n)} -f dA =$$_ D(n) f dA $$, we get$$$$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$$$(d)If$$ba$$and$$ $$−b\leq a$$ then$$|b|\leq a.$$ From (a), we got $\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$, From (b) and (c), we got $- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$ Therefore, we can conclude that $\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$

9.

-4.21. Find the point on the plane $z = x − 2y + 3$ that is closest to the origin, by finding where the square of the distance between $$(0,0)$$ and a point $$(x,y)$$ of the plane is at a minimum. Use the matrix of second partial derivatives to show that the point is a local minimum.

Let $D=d^2 = f(x,y)= x^2+y^2+(x-2y+3)^2$, to find the local extrema we let $\triangledown f = (4x−4y+6,−4x+10y−12)=0$ at $$(− 0.5,1).$$ so $H(-0.5,1)= \left[ \begin{array}{ c c } 4 & -4 \\ -4 & 10 \end{array} \right]$ Because 4 > 0 and (4)(10) − (−4)2 = 24 > 0. So by the Theorem 4.3, it is positive definite. By theorem 4.8, If $$\triangledown f (A) = 0$$ and the Hessian matrix $$[f_{x_i x_j} (A])$$ is positive definite at $$A$$, then $$f (A)$$ is a local minimum. Therefore, $$f$$ has a local minimum at point $$(-0.5,1)$$

10.

-7.32. Let $$S$$ be the unit sphere centered at the origin in $$R^3$$. Evaluate the following items, using as little calculation as possible

(a)$$\int_{S} 1 d\sigma$$

(b)$$\int_{S} ||X||^2 d\sigma$$

(c) Verify that $$\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma =\int_{S} x_3^2 d\sigma$$ using either a symmetric argument or parametrizations. Can you do this without evaluating them?

(d) Use the result of parts (b) and (c) to deduce the value of $$\int_{S} x_1^2 d\sigma$$

(a) In geometry, $$\int_{S} 1 d\sigma$$ means the area of the unit sphere in $$R ^3$$ So $$\int_{S} 1 d\sigma= \pi·1^3 =4\pi$$

(b) For all X S we have $$||X||^2 =1$$, therefore $$\int_{S} ||X||^2 d\sigma=\int_{S} 1 d\sigma=4\pi$$

(c) Rotation by /2 about the $$x_3$$-axis corresponds to some transformation on the domain of the parametrization of $$S$$. We know that $$x_1$$ comes to the same position as $$x_2$$, Therefore $$\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma$$ In the same way, make a rotation by $$\pi/2$$ about the $$x_2$$ , we got $$\int_{S} x_1^2 d\sigma = \int_{S} x_3^2 d\sigma$$ Therefore, $$\int_{S} x_1^2 d\sigma = \int_{S} x_2^2 d\sigma =\int_{S} x_3^2 d\sigma$$

(d) By the definition of norm $$||X||$$, we know that $$||X||=x_1^2 +x_2^2 +x_3^2$$ So, $\int_{S} ||X||^2 d\sigma= \int_{S} x_1^2 +x_2^2 +x_3^2 d\sigma= 3\int_{S} x_1^2 d\sigma = 4\pi$ Therefore,$\int_{S} x_1^2 d\sigma =\frac{1}{3}\int_{S} ||X||^2 d\sigma = \frac{4\pi}{3}$